One of the questions from my homework was to find exact lower bound of
(#black nodes)/(#red nodes)
in rb-tree. the bound must be not asymptotic.
Any suggestions?
Your help would be very appreciated.
Assuming this is a homework:
Let's review some properties of RedBlack Trees from Wikipedia:
...
The root is black.
All leaves are black.
Both children of every red node are black.
...
To get a lower bound on #B/#R you want to construct a tree that has as many red nodes as possible. (Unfortunately, due to 2,3,4 you cannot construct an all red tree)
Some questions worth thinking about:
can you fit more red nodes in balanced or not-so-balanced trees?
does even or odd maximal height make a difference?
given that a tree contains 3, 7, ..., (2^n)-1 back nodes how many red ones can you fit in?
Related
I've been learning and working on implementing a red-black tree data structure. I'm following this article on red-black tree deletion examples and looking at example 5 they have:
When I insert the same nodes into my tree, I get the following:
I understand that red black trees are not unique (I think), therefore both of the above trees are valid since they don't violate any of the properties.
In the example article, after deleting node 1, they get the following:
But after deleting node 1 in my code, I get the following:
Since in my case, node 1 is red, I don't call my delete_fix function which takes care of re-arranging the tree and such. The deletion algorithm I was following simply states to call a delete_fix function if the node to be deleted is black.
However, after comparing my tree with the one in the example article I can see that mine is not exactly optimized. It still follows the rules of the red-black tree though. Is this to be expected with red-black trees or am I missing something here?
However, after comparing my tree with the one in the example article I can see that mine is not exactly optimized.
It is optimised. Your tree will be fast at deleting nodes 5, 7, 20 & 28. The other only 5 & 7.
Bear in mind that for Red-Black Trees, they can be bushy in one direction. If the black tree height of real nodes is N, then the minimum path from root to leaf node is N (all black) and maximum path from root to leaf node is 2 * N (alternatively black-red-black-red etc). If you try to add a new node to the bushy path that is at maximum height, the tree will recolour and/or rebalance.
If you want a more balanced search tree you should use an AVL tree. Red-Black trees favour minimal insertion/deletion fixups over finding a node. Your tree is fine.
Properties of Red-Black Tree:
Every node is either red or black.
The root is black.
Every leaf (NIL) is black.
If a node is red, then both its children are black.
For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.
According to the properties, are these valid or invalid red black trees?
A.
I think this is valid
B.
I think this is valid, but I am not sure since there two adjacent red nodes?
C.
I think this is valid, but I am not sure since there two adjacent red nodes?
D.
I think this is not valid since it violate Property 4?
Did I understand these properties of a RBtree right? If not, where am I wrong?
You have listed the properties of Red-Black trees correctly. Of the four trees only C is not a valid red-black tree:
A.
This is a valid tree. Wikipedia confirms:
every perfect binary tree that consists only of black nodes is a red–black tree.
B.
I think this is valid, but I am not sure since there two adjacent red nodes?
It is valid. There is no problem with red nodes being siblings. They just should not be in a parent-child relationship.
C.
I think this is valid, but I am not sure since there two adjacent red nodes?
It is not valid. Not because of the adjacent red nodes, but because of property 5. The node with label 12 has paths to its leaves with varying number of black nodes. Same for the node 25.
As a general rule, a red node can never have exactly one NIL-leaf as child. Its children should either both be NIL-leaves, or both be (black) internal nodes. This follows from the properties.
D.
I think this is not valid since it violate Property 4?
Property 4 is not violated: the children of the red nodes are NIL leaves (not visualised here), which are black. The fact that these red nodes have black NIL leaves as siblings is irrelevant: there are no rules that concern siblings. So this is valid.
For an example that combines characteristics of tree C and D, see this valid tree depicted in the Wikipedia article, which also depicts the NIL leaves:
A, B & D are valid red-black trees
C is not valid red-black tree as the black height from root to leaf is not the same. It is 2 in some paths and 1 in other paths. It violates what you stated as rule 5.
If 12 had a right child that was black and 25 a left child that was black, then it would be a red-black tree.
A red-black tree is basically identical to a 2-3-4 tree(4-Btree), even though the splitting/swapping method is upside down.
2-3-4 trees have fixed-size 3-node buckets. The color black means that it's the central node of the 3-bucket. Any red-black tree is considered as a perfect quadtree/binary tree (of 3-node-buckets) with empty nodes(black holes and red holes).
In other words, every black node (every 3-bucket) has its absolute position in the perfect tree(2 dimensional unique Cartesian or 4-adic/2-adic unique fraction number).
NIL nodes are just extra flags to save space; you don't have enough memory to store a perfect quadtree/binary tree.
The easiest way to check a red-black tree is to check that each black node is a new bucket(going down) and each red node is grouped with the above black node(same bucket). If the central black node has less than 2 red nodes, you can just add empty red holes next to the central black node(left and right).
A new black node is always the grandson of the last black node, and each black node can have only two red daughter-nodes and no black son-nodes. If the red daughter(mother) is empty(dead/unborn), the motherless grandson-node is directly linked to its grandfather-node.
A motherless black grandson-node has no brother, but he can have a black cousin-node next to him; the 2 cousins are linked to the same grandfather.
A quadtree is a subset of a binary tree.
All black nodes have even heights(2,4,6...), and all red nodes have odd heights(1,3,5...). Optionally, you can use the half unit 0.5.
The 3-bucket has a fixed size 3; just add extra red holes(unborn unlinked red daughters) to make the size 3.
The smallest number of internal nodes in a red-black tree with black height of k is 2k-1 which is one in the following image:
The largest number of internal nodes with black height of k is 22k-1 which, if the black height is 2, should be 24 - 1 = 15. However, consider this image:
The number of internal nodes is 7. What am I doing wrong?
(I've completely rewritten this answer because, as the commenters noted, it was initially incorrect.)
I think it might help to think about this problem by using the isometry between red-black trees and 2-3-4 trees. Specifically, a red-black tree with black height h corresponds to a 2-3-4 tree with height h, where each red node corresponds to a key in a multi-key node.
This connection makes it easier for us to make a few neat observations. First, any 2-3-4 tree node in the bottom layer corresponds to a black node with either no red children, one red child, or two red children. These are the only nodes that can be leaf nodes in the red-black tree. If we wanted to maximize the number of total nodes in the tree, we'd want to make the 2-3-4 tree have nothing but 4-nodes, which (under the isometry) maps to a red/black tree where every black node has two red children. An interesting effect of this is that it makes the tree layer colors alternate between black and red, with the top layer (containing the root) being black.
Essentially, this boils down to counting the number of internal nodes in a complete binary tree of height 2h - 1 (2h layers alternating between black and red). This is equal to the number of nodes in a complete binary tree of height 2h - 2 (since if you pull off all the leaves, you're left with a complete tree of height one less than what you started with). This works out to 22h - 1 - 1, which differs from the number that you were given (which I'm now convinced is incorrect) but matches the number that you're getting.
You need to count the black NIL leafs in the tree if not this formula won't work. The root must not be RED that is in violation of one of the properties of a Red-Black tree.
The problem is you misunderstood the black height.
The black height of a node in a red-black tree is the the number of black nodes from the current node to a leaf not counting the current node. (This will be the same value in every route).
So if you just add two black leafs to every red node you will get a red-black tree with a black height of 2 and 15 internal nodes.
(Also in a red-black tree every red node has two black children so red nodes can't be leafs.)
After reading the discussion above,so if I add the root with red attribute, the second node I add will be a red again which would be a red violation, and after node restructuring, I assume that we again reach root black and child red ! with which we might not get (2^2k)-1 max internal nodes.
Am I missing something here , started working on rbt just recently ...
It seems you havent considered the "Black Leaves" (Black nodes) -- the 2 NIL nodes for each of the Red Nodes on the last level. If you consider the NIL nodes as leaves, the Red nodes on the last level now get counted as internal nodes totaling to 15.
The tree given here actually has 15 internal nodes. The NIL black children of red nodes in last layer are missing which are actually called external nodes ( node without a key ). The tree has black-height of 2. The actual expression for maximum number of internal nodes for a tree with black-height k is 4^(k)-1. In this case, it turns out to be 15.
In red-black trees, external nodes[null nodes] are always black but in your question for the second tree you have not mentioned external nodes and hence you are getting your count as 7 but if u mention external nodes[null nodes] and then count internal nodes you can see that it turns out to be 15.
Not sure that i understand the question.
For any binary tree where all layers (except maybe last one) have max number of items we will have 2^(k-1)-1 internal nodes, where k is number of layers. At second picture you have 4 layers, so number of internal nodes is 2^(4-1)-1=7
I have to find a family of maximally unbalanced red-black trees and to prove the "respective attributes" of that family to prove that there is an infinitely big family of red black trees that have a height close to 2log(n+1).
Now my guess is that this family consists of basically all the red black trees that have one path with s-r-s-r ... nodes and the rest filled with black nodes. But how do I prove this? and how do i formally write down how such a family looks like?
Thank you!
Now my guess is that this family consists of basically all the red black trees that have one path with s-r-s-r ... nodes and the rest filled with black nodes.
That's a reasonable guess.
But how do I prove this?
Describe an infinite sequence of trees T_0, T_1, T_2, T_3, ..., such that, for every integer n, there exists a tree in the sequence with at least n nodes. Show that there exists a constant C such that, for every i, the height of T_i is at least 2log(n_i+1) - C, where n_i is the number of nodes in T_i. (This is one possible interpretation of the ambiguous term "close to".)
how do i formally write down how such a family looks like?
Inductively. I'll do the all-black trees as an example. The tree T_0 is empty (base case). For all integers i > 0, the tree T_i consists of a black node with left and right subtrees equal to T_{i-1} (inductive step). Then you can prove facts about these trees using induction.
I have an exam next week in algorithms, and was given questions to prepare for it. One of these questions has me stumped though.
"Can we draw a red-black tree with 7 black nodes and 10 red nodes? why?"
It sounds like it could be answered quickly, but I can't get my mind around it.
The CRLS gives us the maximum height of a RB tree with n internal nodes: 2*lg(n+1).
I think the problem could be solved using this lemma alone, but am not sure.
Any tips?
Since this is exam preparation, I don't want to give you a direct answer, but I think what you need to consider is the properties that govern how you build a Red-Black Tree:
A node is either red or black.
The root is black. (This rule is sometimes omitted from other definitions. Since the root can always be changed from red to black but not necessarily vice-versa this rule has little effect on analysis.)
All leaves are black.
Both children of every red node are black.
Every simple path from a given node to any of its descendant leaves contains the same number of black nodes.
(Stole these from the wikipedia page: http://en.wikipedia.org/wiki/Red-black_tree)
Given the count of nodes you listed, can you meet all of these properties?
the answer is simple.
As we know that a red node can have only black parent.The max no. of nodes will be when each black node's both children are red and, hence, every black node has red parent. So, for 'n' black nodes '2n' red node are possible.
Think it this way:
put the first node(which is root) & make it black
make both its children red
make left and right children of both these red nodes black and for all these black nodes,
follow the same procedure as followed with root until black node count reaches the given value (in this case 7)
hope this helped you visualize the solution.
The answer critically depends on whether your RB tree uses black dummy nodes at the leaves, and if so, they are included in the count of seven black nodes. If not, consider a complete tree of seven black nodes
*
/ \
* *
/\ /\
* * * *
You won't have much trouble adding ten red nodes.